What distinguishes the sample standard deviation formula from the population version?
Correct Answer
D) The sample version divides by one less than the count
The usual sample variance divides the sum of squared deviations from the sample mean by n−1, and sample SD is its square root. The corresponding population-form variance divides by N. Both square deviations about a mean and include all observations. Bessel’s correction makes sample variance unbiased under standard sampling assumptions; the square-root sample SD is not thereby an unbiased estimator of population SD.
Why This Is the Correct Answer
Option D is correct because the usual sample standard deviation divides the sum of squared deviations from the sample mean by one less than the count (n − 1), then takes the square root. The population version divides by N. This n − 1 divisor, known as Bessel's correction, makes the sample variance unbiased under standard sampling assumptions.
Why the Other Options Are Wrong
Option A: The sample version omits the squaring of deviations
Option A says the sample version omits the squaring of deviations. Both versions square each deviation from the mean; omitting the squaring would make positive and negative deviations cancel.
Option B: The sample version uses the median instead of a mean
Option B says the sample version uses the median instead of a mean. Both the sample and population formulas measure deviations from a mean, not from the median.
Option C: The sample version discards the highest observation
Option C claims the sample version discards the highest observation. Both formulas include all observations; the difference lies only in the divisor, n − 1 versus N.
Sample loses one
A sample gives up one degree of freedom because its mean was estimated from the same data, so the sample formula divides by n − 1.
How to use: Use this to recall which divisor belongs to the sample standard deviation.
Exam Tip
Remember that the only formula difference is the divisor; options that change squaring, the mean or the data set are wrong.
Common Mistakes to Avoid
- -Dividing by n when the question treats the data as a sample
- -Believing the sample formula drops extreme observations
- -Assuming the sample standard deviation is an unbiased estimate of the population standard deviation
Concept Deep Dive
Analysis
This question tests the difference between the sample and population standard deviation formulas. Both square the deviations from the mean and include all observations. The sample version divides the sum of squared deviations by n − 1 rather than N before taking the square root.
Background Knowledge
Standard deviation measures how spread out values are around the mean. When working from a sample, dividing by n − 1 corrects the tendency of the sample variance to underestimate the population variance. The square root of the unbiased sample variance is not itself an unbiased estimator of the population standard deviation.
Real-World Application
An appraiser computes the standard deviation of 12 adjusted comparable prices treated as a sample of the market. Using n − 1 = 11 as the divisor produces a slightly larger figure than dividing by 12, reflecting the uncertainty of estimating the mean from the same sales.
More Statistics Questions
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Previous Question
A sale-price distribution has fourth central moment m4 = 196,830,000 in (thousands of dollars)^4 and second central moment m2 = 8100 in (thousands of dollars)^2. Compute Pearson moment kurtosis m4 / m2². Use moments with the same population (N) divisor, not an unbiased sample variance; report two decimals.
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An appraiser assigns reliability weights of 2, 3, 1 and 2 to sales at $195,000, $240,000, $270,000 and $330,000, respectively. What is the weighted mean price, to the nearest dollar?